1 Data

Information about the test:

question description
A1 linear function
A2 3D
A3 arithmetic rules
A4 Easy ineq.
A5 logs
A6 logs
A7 graph translation
A8 sine pi/3
A9 trig.ineq.
A10 trig.identity
A11 period
A12 rational exponents
A13 ellipsoid
A14 limit
A15 series
A16 diff.quotient
A17 graph f'
A18 product rule
A19 quotient rule
A20 (exp)'
A21 (ln(sin))'
A22 hyp.functions
A23 slope tangent
A24 IVT
A25 velocity
A26 int(poly)
A27 int(exp)
A28 Int = 0
A29 int even funct
A31 int(abs)
A32 FtoC algebra
A33 difference vectors
A35 intersect planes
A36 parallel planes
A30 FtoC graph
A34 normal vector

Load the student scores for the test - here we load the 2017 and 2018 ETH Zurich test data:

test_scores
## # A tibble: 3,433 × 38
##    year  class    A1    A2    A3    A4    A5    A6    A7    A8    A9   A10   A11
##    <chr> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
##  1 2017  s21t…     1     0     1     1     1     0     1     0     1     0     1
##  2 2017  s21t…     1     0     1     1     1     1     0     1     1     2     1
##  3 2017  s21t…     1     0     0     0     1     1     1     0     1     1     1
##  4 2017  s21t…     1     0     1     1     1     1     1     1     1     1     1
##  5 2017  s21t…     1     0     1     0     2     0     1     0     1     0     2
##  6 2017  s21t…     0     1     0     0     1     2     0     2     2     2     1
##  7 2017  s21t…     1     0     1     0     2     1     0     1     2     1     0
##  8 2017  s21t…     1     1     1     1     2     1     1     2     2     2     2
##  9 2017  s21t…     1     1     0     1     1     1     1     1     1     1     1
## 10 2017  s21t…     1     0     1     0     0     1     0     0     1     1     1
## # … with 3,423 more rows, and 25 more variables: A12 <dbl>, A13 <dbl>,
## #   A14 <dbl>, A15 <dbl>, A16 <dbl>, A17 <dbl>, A18 <dbl>, A19 <dbl>,
## #   A20 <dbl>, A21 <dbl>, A22 <dbl>, A23 <dbl>, A24 <dbl>, A25 <dbl>,
## #   A26 <dbl>, A27 <dbl>, A28 <dbl>, A29 <dbl>, A30 <dbl>, A31 <dbl>,
## #   A32 <dbl>, A33 <dbl>, A34 <dbl>, A35 <dbl>, A36 <dbl>

1.1 Missing data

The data includes scores of 2 for the “I don’t know” answer option. We replace these with 0, reflecting a non-correct answer:

test_scores <- test_scores %>% 
  mutate(across(starts_with("A"), ~ ifelse(. == 2, 0, .)))

1.2 Data summary

The number of responses from each class:

test_scores <- test_scores %>% 
  rowwise() %>% 
  mutate(Total = rowSums(across(starts_with("A"))))

test_scores_summary <- test_scores %>% 
  group_by(year) %>% 
  summarise(
    n = n(),
    mean = mean(Total),
    sd = sd(Total),
    median = median(Total)
  )

test_scores_summary %>% 
  gt() %>% 
  fmt_number(columns = c("mean", "sd"), decimals = 2) %>%
  data_color(
    columns = c("n"),
    colors = scales::col_numeric(palette = c("Blues"), domain = NULL)
  )
year n mean sd median
2017 1682 18.68 8.44 19
2018 1751 19.03 8.20 19
p1 <- test_scores %>% 
  ggplot(aes(x = Total)) +
  geom_histogram(binwidth = 2) +
  #scale_x_continuous(limits = c(0,100), breaks = c(0, 50, 100)) +
  facet_grid(cols = vars(year)) +
  theme_minimal() +
  labs(x = "Total score (out of 36)",
       y = "Number of students",
       title = "ETH s21t") +
  theme(axis.title.x = element_blank(), axis.title.y = element_blank())

p2 <- test_scores_summary %>% 
  mutate(
    n = str_glue("{n}"),
    mean = str_glue("{round(mean, digits = 1)}"),
    sd = str_glue("{round(sd, digits = 1)}"),
    median = str_glue("{median}")
  ) %>% 
  pivot_longer(c(n, mean, sd, median), names_to = "layer", values_to = "label") %>% 
  mutate(layer = fct_relevel(layer, c("n", "sd", "mean", "median")) %>% fct_rev()) %>% 
  ggplot(aes(x = 80, y = layer, label = label)) +
  geom_text(size = 10 * 5/14, hjust = 1) +
  scale_x_continuous(limits = c(0,100)) +
  facet_grid(cols = vars(year)) +
  labs(y = "", x = NULL) +
  scale_y_discrete(labels = c("n" = "N", "mean" = "Mean", "median" = "Median")) +
  theme_minimal() +
  theme(axis.line = element_blank(), axis.ticks = element_blank(), axis.text.x = element_blank(),
        panel.grid = element_blank(), strip.text = element_blank())

p1 / p2 +  plot_layout(heights = c(5, 2.5))

ggsave("output/eth_pre_data-summary.pdf", units = "cm", width = 8, height = 8)

Mean and standard deviation for each item:

test_scores %>% 
  select(-class, -Total) %>% 
  group_by(year) %>% 
  skim_without_charts() %>% 
  select(-contains("character."), -contains("numeric.p"), -skim_type) %>% 
  rename(complete = complete_rate) %>% 
  # make the table wider, i.e. with separate columns for each year's results, with the year at the start of each column name
  pivot_wider(names_from = year, values_from = -c(skim_variable, year), names_glue = "{year}__{.value}") %>% 
  # put the columns in order by year
  select(sort(names(.))) %>% 
  select(skim_variable, everything()) %>% 
  # use GT to make the table look nice
  gt(rowname_col = "skim_variable") %>% 
  # group the columns from each year
  tab_spanner_delim(delim = "__") %>%
  fmt_number(columns = contains("numeric"), decimals = 2) %>%
  fmt_percent(columns = contains("complete"), decimals = 0) %>% 
  # change all the numeric.mean and numeric.sd column names to Mean and SD
  cols_label(
    .list = test_scores %>% select(year) %>% distinct() %>% transmute(col = paste0(year, "__numeric.mean"), label = "Mean") %>% deframe()
  ) %>% 
  cols_label(
    .list = test_scores %>% select(year) %>% distinct() %>% transmute(col = paste0(year, "__numeric.sd"), label = "SD") %>% deframe()
  ) %>%
  data_color(
    columns = contains("numeric.mean"),
    colors = scales::col_numeric(palette = c("Greens"), domain = NULL)
  )
2017 2018
complete n_missing Mean SD complete n_missing Mean SD
A1 100% 0 0.95 0.22 100% 0 0.95 0.21
A2 100% 0 0.31 0.46 100% 0 0.28 0.45
A3 100% 0 0.62 0.48 100% 0 0.64 0.48
A4 100% 0 0.64 0.48 100% 0 0.62 0.48
A5 100% 0 0.47 0.50 100% 0 0.49 0.50
A6 100% 0 0.71 0.46 100% 0 0.73 0.44
A7 100% 0 0.71 0.46 100% 0 0.70 0.46
A8 100% 0 0.47 0.50 100% 0 0.50 0.50
A9 100% 0 0.46 0.50 100% 0 0.46 0.50
A10 100% 0 0.54 0.50 100% 0 0.54 0.50
A11 100% 0 0.57 0.49 100% 0 0.58 0.49
A12 100% 0 0.68 0.47 100% 0 0.71 0.45
A13 100% 0 0.37 0.48 100% 0 0.36 0.48
A14 100% 0 0.56 0.50 100% 0 0.57 0.49
A15 100% 0 0.46 0.50 100% 0 0.46 0.50
A16 100% 0 0.19 0.39 100% 0 0.20 0.40
A17 100% 0 0.59 0.49 100% 0 0.60 0.49
A18 100% 0 0.38 0.49 100% 0 0.36 0.48
A19 100% 0 0.35 0.48 100% 0 0.34 0.47
A20 100% 0 0.72 0.45 100% 0 0.72 0.45
A21 100% 0 0.51 0.50 100% 0 0.52 0.50
A22 100% 0 0.52 0.50 100% 0 0.52 0.50
A23 100% 0 0.41 0.49 100% 0 0.43 0.50
A24 100% 0 0.59 0.49 100% 0 0.58 0.49
A25 100% 0 0.54 0.50 100% 0 0.73 0.44
A26 100% 0 0.80 0.40 100% 0 0.80 0.40
A27 100% 0 0.39 0.49 100% 0 0.40 0.49
A28 100% 0 0.42 0.49 100% 0 0.46 0.50
A29 100% 0 0.48 0.50 100% 0 0.52 0.50
A30 100% 0 0.76 0.43 100% 0 0.73 0.44
A31 100% 0 0.38 0.48 100% 0 0.37 0.48
A32 100% 0 0.23 0.42 100% 0 0.20 0.40
A33 100% 0 0.78 0.42 100% 0 0.76 0.43
A34 100% 0 0.58 0.49 100% 0 0.61 0.49
A35 100% 0 0.32 0.47 100% 0 0.33 0.47
A36 100% 0 0.22 0.41 100% 0 0.23 0.42

2 Testing assumptions

Before applying IRT, we should check that the data satisfies the assumptions needed by the model. In particular, to use a 1-dimensional IRT model, we should have some evidence of unidimensionality in the test scores.

2.1 Inter-item correlations

If the test is unidimensional then we would expect student scores on pairs of items to be correlated.

This plot shows the correlations between scores on each pair of items:

item_scores <- test_scores %>% 
  select(-class, -year, -Total)

cor_ci <- psych::corCi(item_scores, plot = FALSE)

psych::cor.plot.upperLowerCi(cor_ci)

Almost all correlations are significantly different from 0 - the only one that is not is:

cor_ci$ci %>% 
  as_tibble(rownames = "corr") %>% 
  filter(p > 0.05) %>% 
  arrange(-p) %>% 
  select(-contains(".e")) %>% 
  gt() %>% 
  fmt_number(columns = 2:4, decimals = 3)
corr lower upper p
A1-A2 −0.007 0.049 0.134

Thus, the overall picture is that the item scores are well correlated with each other.

2.2 Dimensionality

We use factor analysis as another way to assess unidimensionality. We are looking to see whether a 1-factor solution is feasible.

structure <- check_factorstructure(item_scores)
n <- n_factors(item_scores)

2.2.1 Is the data suitable for Factor Analysis?

  • KMO: The Kaiser, Meyer, Olkin (KMO) measure of sampling adequacy suggests that data seems appropriate for factor analysis (KMO = 0.95).
  • Sphericity: Bartlett’s test of sphericity suggests that there is sufficient significant correlation in the data for factor analysis (Chisq(630) = 30319.01, p < .001).

2.2.1.1 Method Agreement Procedure:

The choice of 6 dimensions is supported by 4 (21.05%) methods out of 19 (Optimal coordinates, Parallel analysis, Kaiser criterion, BIC).

plot(n)

summary(n) %>% gt()
n_Factors n_Methods
1 3
2 1
3 1
4 1
6 4
8 1
11 3
33 1
34 1
35 3
#n %>% tibble() %>% gt()
fa.parallel(item_scores, fa = "fa")

## Parallel analysis suggests that the number of factors =  9  and the number of components =  NA
pdf(file = "output/eth_pre_scree.pdf", width = 6, height = 4)
fa.parallel(item_scores, fa = "fa")
## Parallel analysis suggests that the number of factors =  10  and the number of components =  NA
dev.off()
## png 
##   2

This shows that both the 6-factor and 1-factor solutions are worth exploring.

2.2.2 1 Factor

We use the factanal function to fit a 1-factor model.

Note that this function cannot handle missing data, so any NA scores must be set to 0 for this analysis.

Show factanal output
fitfact <- factanal(item_scores,
                    factors = 1,
                    rotation = "varimax")
print(fitfact, digits = 2, cutoff = 0.3, sort = TRUE)
## 
## Call:
## factanal(x = item_scores, factors = 1, rotation = "varimax")
## 
## Uniquenesses:
##   A1   A2   A3   A4   A5   A6   A7   A8   A9  A10  A11  A12  A13  A14  A15  A16 
## 0.95 0.95 0.75 0.79 0.72 0.74 0.79 0.79 0.61 0.65 0.67 0.88 0.91 0.73 0.87 0.81 
##  A17  A18  A19  A20  A21  A22  A23  A24  A25  A26  A27  A28  A29  A30  A31  A32 
## 0.78 0.78 0.78 0.63 0.59 0.61 0.64 0.73 0.94 0.69 0.63 0.64 0.85 0.81 0.78 0.78 
##  A33  A34  A35  A36 
## 0.88 0.86 0.79 0.84 
## 
## Loadings:
##  [1] 0.53 0.51 0.62 0.59 0.57 0.52 0.61 0.64 0.62 0.60 0.52 0.56 0.61 0.60     
## [16]      0.50 0.45 0.46 0.46 0.35 0.31 0.36 0.44 0.47 0.46 0.47      0.39 0.44
## [31] 0.46 0.47 0.34 0.38 0.46 0.40
## 
##                Factor1
## SS loadings       8.35
## Proportion Var    0.23
## 
## Test of the hypothesis that 1 factor is sufficient.
## The chi square statistic is 5447.17 on 594 degrees of freedom.
## The p-value is 0
load <- tidy(fitfact)

ggplot(load, aes(x = fl1, y = 0)) + 
  geom_point() + 
  geom_label_repel(aes(label = paste0("A", rownames(load))), show.legend = FALSE) +
  labs(x = "Factor 1", y = NULL,
       title = "Standardised Loadings", 
       subtitle = "Based upon correlation matrix") +
  theme_minimal()
## Warning: ggrepel: 14 unlabeled data points (too many overlaps). Consider
## increasing max.overlaps

load %>% 
  select(question = variable, factor_loading = fl1) %>% 
  left_join(item_info, by = "question") %>% 
  arrange(-factor_loading) %>% 
  gt() %>%
  data_color(
    columns = contains("factor"),
    colors = scales::col_numeric(palette = c("Greens"), domain = NULL)
  )
question factor_loading description
A21 0.6407106 (ln(sin))'
A9 0.6228304 trig.ineq.
A22 0.6223752 hyp.functions
A27 0.6076260 int(exp)
A20 0.6061690 (exp)'
A23 0.5998775 slope tangent
A28 0.5961137 Int = 0
A10 0.5878608 trig.identity
A11 0.5709488 period
A26 0.5590386 int(poly)
A5 0.5324571 logs
A14 0.5232019 limit
A24 0.5215894 IVT
A6 0.5086087 logs
A3 0.4972984 arithmetic rules
A19 0.4718772 quotient rule
A17 0.4705175 graph f'
A32 0.4683131 FtoC algebra
A18 0.4646006 product rule
A31 0.4643091 int(abs)
A8 0.4602708 sine pi/3
A7 0.4593494 graph translation
A35 0.4579597 intersect planes
A4 0.4547066 Easy ineq.
A30 0.4376910 FtoC graph
A16 0.4367742 diff.quotient
A36 0.4014553 parallel planes
A29 0.3875447 int even funct
A34 0.3765420 normal vector
A15 0.3603866 series
A12 0.3513586 rational exponents
A33 0.3415341 difference vectors
A13 0.3075858 ellipsoid
A25 0.2364578 velocity
A2 0.2342048 3D
A1 0.2277957 linear function

This factor appears to be dominated by “procedural calculus skills”: standard calculation of derivatives, integral and some trigonometry.

2.2.3 6 Factor

Here we also investigate the 6-factor solution, to see whether these factors are interpretable.

Show factanal output
fitfact6 <- factanal(item_scores, factors = 6, rotation = "varimax")
print(fitfact6, digits = 2, cutoff = 0.3, sort = TRUE)
## 
## Call:
## factanal(x = item_scores, factors = 6, rotation = "varimax")
## 
## Uniquenesses:
##   A1   A2   A3   A4   A5   A6   A7   A8   A9  A10  A11  A12  A13  A14  A15  A16 
## 0.91 0.81 0.74 0.75 0.68 0.74 0.72 0.76 0.53 0.64 0.64 0.87 0.80 0.71 0.82 0.72 
##  A17  A18  A19  A20  A21  A22  A23  A24  A25  A26  A27  A28  A29  A30  A31  A32 
## 0.69 0.41 0.00 0.47 0.53 0.59 0.60 0.63 0.91 0.58 0.60 0.61 0.83 0.71 0.76 0.75 
##  A33  A34  A35  A36 
## 0.84 0.70 0.56 0.70 
## 
## Loadings:
##     Factor1 Factor2 Factor3 Factor4 Factor5 Factor6
## A9   0.54    0.37                                  
## A23  0.50                                          
## A28  0.50                                          
## A24          0.52                                  
## A20                  0.59                          
## A18                          0.69                  
## A19                          0.95                  
## A35                                  0.58          
## A1                                                 
## A2                                           0.40  
## A3   0.31                                          
## A4           0.33                                  
## A5   0.38                                          
## A6                                                 
## A7           0.44                                  
## A8   0.39                                          
## A10  0.45                                          
## A11  0.43    0.36                                  
## A12                                                
## A13                                          0.34  
## A14  0.32                                          
## A15                                                
## A16  0.43                                          
## A17          0.48                                  
## A21  0.46            0.43                          
## A22  0.41            0.36                          
## A25                                                
## A26          0.34    0.48                          
## A27  0.45            0.32                          
## A29                                                
## A30          0.48                                  
## A31  0.33                                          
## A32  0.41                                          
## A33          0.33                                  
## A34                                  0.47          
## A36                                  0.43          
## 
##                Factor1 Factor2 Factor3 Factor4 Factor5 Factor6
## SS loadings        3.5    2.65    1.68    1.62    1.26    0.98
## Proportion Var     0.1    0.07    0.05    0.05    0.04    0.03
## Cumulative Var     0.1    0.17    0.22    0.26    0.30    0.32
## 
## Test of the hypothesis that 6 factors are sufficient.
## The chi square statistic is 932.36 on 429 degrees of freedom.
## The p-value is 2.36e-39
load6 <- tidy(fitfact6)

ggplot(load6, aes(x = fl1, y = fl2)) + 
  geom_point() + 
  geom_label_repel(aes(label = paste0("A", rownames(load))), show.legend = FALSE) +
  labs(x = "Factor 1", y = "Factor 2",
       title = "Standardised Loadings", 
       subtitle = "Based upon correlation matrix") +
  theme_minimal()

main_factors <- load6 %>% 
#  mutate(factorNone = 0.4) %>%  # add this to set the main factor to "None" where all loadings are below 0.4
  pivot_longer(names_to = "factor",
               cols = contains("fl")) %>% 
  mutate(value_abs = abs(value)) %>% 
  group_by(variable) %>% 
  top_n(1, value_abs) %>% 
  ungroup() %>% 
  transmute(main_factor = factor, variable)

load6 %>% 
  select(-uniqueness) %>% 
  # add the info about which is the main factor
  left_join(main_factors, by = "variable") %>%
  left_join(item_info %>% select(variable = question, description), by = "variable") %>% 
  arrange(main_factor) %>% 
  select(main_factor, everything()) %>% 
  # arrange adjectives by descending loading on main factor
  rowwise() %>% 
  mutate(max_loading = max(abs(c_across(starts_with("fl"))))) %>% 
  group_by(main_factor) %>% 
  arrange(-max_loading, .by_group = TRUE) %>% 
  select(-max_loading) %>% 
  # sort out the presentation
  rename("Main Factor" = main_factor, # the _ throws a latex error
         "Question" = variable) %>%
  mutate_at(
    vars(starts_with("fl")),
    ~ cell_spec(round(., digits = 3), bold = if_else(abs(.) > 0.4, T, F))
  ) %>% 
  kable(booktabs = T, escape = F, longtable = T) %>% 
  kableExtra::collapse_rows(columns = 1, valign = "top") %>%
  kableExtra::kable_styling(latex_options = c("repeat_header"))
Main Factor Question fl1 fl2 fl3 fl4 fl5 fl6 description
fl1 A9 0.543 0.371 0.045 0.1 0.13 0.086 trig.ineq.
fl1 A23 0.504 0.289 0.189 0.085 0.136 0.023 slope tangent
fl1 A28 0.503 0.292 0.145 0.094 0.147 0.048 Int = 0
fl1 A21 0.456 0.194 0.429 0.13 0.143 0.07 (ln(sin))’
fl1 A27 0.453 0.193 0.316 0.086 0.216 0.075 int(exp)
fl1 A10 0.447 0.239 0.234 0.104 0.084 0.182 trig.identity
fl1 A16 0.427 -0.017 0.141 0.129 0.128 0.217 diff.quotient
fl1 A11 0.426 0.357 0.127 0.065 0.082 0.15 period
fl1 A22 0.408 0.237 0.36 0.108 0.21 0.073 hyp.functions
fl1 A32 0.407 0.122 0.106 0.104 0.19 0.115 FtoC algebra
fl1 A8 0.386 0.145 0.138 0.12 0.058 0.186 sine pi/3
fl1 A5 0.379 0.182 0.2 0.093 0.114 0.278 logs
fl1 A31 0.325 0.262 0.059 0.07 0.178 0.151 int(abs)
fl1 A14 0.324 0.176 0.266 0.118 0.18 0.18 limit
fl1 A3 0.313 0.212 0.247 0.085 0.08 0.213 arithmetic rules
fl1 A6 0.293 0.268 0.257 0.072 0.122 0.136 logs
fl2 A24 0.272 0.516 0.074 0.079 0.134 0.049 IVT
fl2 A30 0.171 0.484 0.086 0.051 0.094 0.08 FtoC graph
fl2 A17 0.193 0.478 0.118 0.073 0.061 0.137 graph f’
fl2 A7 0.167 0.443 0.157 0.036 0.095 0.161 graph translation
fl2 A4 0.225 0.334 0.118 0.064 0.084 0.249 Easy ineq.
fl2 A33 0.101 0.328 0.089 0.036 0.113 0.158 difference vectors
fl2 A1 0.046 0.27 0.089 0.027 0.029 0.06 linear function
fl2 A25 0.056 0.264 0.099 0.009 0.055 0.054 velocity
fl2 A12 0.185 0.191 0.143 0.083 0.072 0.168 rational exponents
fl3 A20 0.286 0.268 0.593 0.09 0.115 0.059 (exp)’
fl3 A26 0.221 0.338 0.476 0.11 0.142 0.002 int(poly)
fl3 A29 0.23 0.126 0.27 0.093 0.145 0.03 int even funct
fl4 A19 0.227 0.071 0.129 0.951 0.101 0.085 quotient rule
fl4 A18 0.211 0.138 0.135 0.692 0.114 0.106 product rule
fl5 A35 0.216 0.133 0.114 0.104 0.583 0.102 intersect planes
fl5 A34 0.109 0.185 0.169 0.057 0.47 0.006 normal vector
fl5 A36 0.223 0.106 0.06 0.045 0.43 0.212 parallel planes
fl6 A2 0.093 0.117 0.011 0.046 0.024 0.403 3D
fl6 A13 0.099 0.255 -0.006 0.044 0.106 0.339 ellipsoid
fl6 A15 0.154 0.233 0.084 0.023 0.143 0.28 series

Here, the first factor seems better described by “abstract understanding of calculus”. The second is based on “graphical understanding of calculus”.

The third and fourth factors are “procedural calculus skills”, with the product and quotient rules distinguished in the fourth factor.

The fifth and sixth factors have a geometrical flavour, with the fifth factor in particular being about “vector geometry”.

2.2.4 Conclusion

The 6-factor solution does seem to be interpretable, however the 1-factor solution is still reasonable. In terms of explaining the variance in the data, the 1-factor solution does a good job at capturing a large chunk:

# Compute the proportion of variance explained; see https://stackoverflow.com/a/58159992/17459126
x1 <- loadings(fitfact)
propvar1 <- colSums(x1^2)/nrow(x1)

rbind(`Proportion Var` = propvar1, `Cumulative Var` = cumsum(propvar1))
##                  Factor1
## Proportion Var 0.2320524
## Cumulative Var 0.2320524
x6 <- loadings(fitfact6)
propvar6 <- colSums(x6^2)/nrow(x6)

rbind(`Proportion Var` = propvar6, `Cumulative Var` = cumsum(propvar6))
##                   Factor1    Factor2    Factor3    Factor4    Factor5
## Proportion Var 0.09717865 0.07370191 0.04679681 0.04500204 0.03500279
## Cumulative Var 0.09717865 0.17088056 0.21767737 0.26267941 0.29768220
##                   Factor6
## Proportion Var 0.02729802
## Cumulative Var 0.32498023

Therefore, we are justified in proceeding with a unidimensional IRT model.

3 Fitting 2 parameter logistic MIRT model

We can fit a Multidimensional Item Response Theory (mirt) model. From the function definition:

mirt fits a maximum likelihood (or maximum a posteriori) factor analysis model to any mixture of dichotomous and polytomous data under the item response theory paradigm using either Cai's (2010) Metropolis-Hastings Robbins-Monro (MHRM) algorithm.

The process is to first fit the model, and save the result as a model object that we can then parse to get tabular or visual displays of the model that we might want. When fitting the model, we have the option to specify a few arguments, which then get interpreted as parameters to be passed to the model.

Show model fitting output
fit_2pl <- mirt(
  data = item_scores, # just the columns with question scores
  model = 1,          # number of factors to extract
  itemtype = "2PL",   # 2 parameter logistic model
  SE = TRUE           # estimate standard errors
  )
## 
Iteration: 1, Log-Lik: -66668.278, Max-Change: 0.71582
Iteration: 2, Log-Lik: -65483.178, Max-Change: 0.26188
Iteration: 3, Log-Lik: -65316.598, Max-Change: 0.14767
Iteration: 4, Log-Lik: -65249.858, Max-Change: 0.11072
Iteration: 5, Log-Lik: -65212.345, Max-Change: 0.09658
Iteration: 6, Log-Lik: -65190.906, Max-Change: 0.06270
Iteration: 7, Log-Lik: -65178.102, Max-Change: 0.06095
Iteration: 8, Log-Lik: -65169.976, Max-Change: 0.03668
Iteration: 9, Log-Lik: -65165.338, Max-Change: 0.03883
Iteration: 10, Log-Lik: -65162.425, Max-Change: 0.02443
Iteration: 11, Log-Lik: -65160.764, Max-Change: 0.01661
Iteration: 12, Log-Lik: -65159.750, Max-Change: 0.01247
Iteration: 13, Log-Lik: -65158.047, Max-Change: 0.00237
Iteration: 14, Log-Lik: -65158.024, Max-Change: 0.00195
Iteration: 15, Log-Lik: -65158.012, Max-Change: 0.00168
Iteration: 16, Log-Lik: -65157.995, Max-Change: 0.00082
Iteration: 17, Log-Lik: -65157.990, Max-Change: 0.00089
Iteration: 18, Log-Lik: -65157.986, Max-Change: 0.00072
Iteration: 19, Log-Lik: -65157.976, Max-Change: 0.00018
Iteration: 20, Log-Lik: -65157.975, Max-Change: 0.00020
Iteration: 21, Log-Lik: -65157.975, Max-Change: 0.00019
Iteration: 22, Log-Lik: -65157.974, Max-Change: 0.00012
Iteration: 23, Log-Lik: -65157.973, Max-Change: 0.00012
Iteration: 24, Log-Lik: -65157.973, Max-Change: 0.00011
Iteration: 25, Log-Lik: -65157.973, Max-Change: 0.00009
## 
## Calculating information matrix...

3.1 Local independence

We compute Yen’s \(Q_3\) (1984) to check for any dependence between items after controlling for \(\theta\). This gives a score for each pair of items, with scores above 0.2 regarded as problematic (see DeMars, p. 48).

residuals_2pl  %>% as.matrix() %>% 
  corrplot::corrplot(type = "upper")

This shows that most item pairs are independent, with only one pair showing cause for concern:

residuals_2pl %>%
  rownames_to_column(var = "q1") %>%
  as_tibble() %>% 
  pivot_longer(cols = starts_with("A"), names_to = "q2", values_to = "Q3_score") %>% 
  filter(abs(Q3_score) > 0.2) %>% 
  filter(parse_number(q1) < parse_number(q2)) %>%
  gt()
q1 q2 Q3_score
A18 A19 0.6740812
A34 A35 0.2109123

Items A18 and A19 are on the product and quotient rules.

Given that this violation of the local independence assumption is very mild, we proceed using this model.

3.2 Model parameters

We then compute factor score estimates and augment the existing data frame with these estimates, to keep everything in one place. To do the estimation, we use the fscores() function from the mirt package which takes in a computed model object and computes factor score estimates according to the method specified. We will use the EAP method for factor score estimation, which is the “expected a-posteriori” method, the default. We specify it explicitly below, but the results would have been the same if we omitted specifying the method argument since it’s the default method the function uses.

test_scores_with_2pl_ability <- test_scores %>%
  bind_cols(fscores(fit_2pl, method = "EAP"))

We can also calculate the model coefficient estimates using a generic function coef() which is used to extract model coefficients from objects returned by modeling functions. We will set the IRTpars argument to TRUE, which means slope intercept parameters will be converted into traditional IRT parameters.

coefs_2pl <- coef(fit_2pl, IRTpars = TRUE)

The resulting object coefs is a list, with one element for each question, and an additional GroupPars element that we won’t be using. The output is a bit long, so we’re only showing a few of the elements here:

coefs_2pl[1:3]
## $A1
##                a         b  g  u
## par     1.219104 -2.946041  0  1
## CI_2.5  1.024917 -3.298279 NA NA
## CI_97.5 1.413291 -2.593803 NA NA
## 
## $A2
##                 a        b  g  u
## par     0.5612745 1.670103  0  1
## CI_2.5  0.4752132 1.405353 NA NA
## CI_97.5 0.6473358 1.934852 NA NA
## 
## $A3
##                a          b  g  u
## par     1.354723 -0.5385656  0  1
## CI_2.5  1.236060 -0.6126183 NA NA
## CI_97.5 1.473386 -0.4645129 NA NA
# coefs_2pl[35:37]

Let’s take a closer look at the first element:

coefs_2pl[1]
## $A1
##                a         b  g  u
## par     1.219104 -2.946041  0  1
## CI_2.5  1.024917 -3.298279 NA NA
## CI_97.5 1.413291 -2.593803 NA NA

In this output:

  • a is discrimination
  • b is difficulty
  • endpoints of the 95% confidence intervals are also shown

To make this output more user friendly, we have produced a function tidy_mirt_coeffs that transforms it into a tidy table.

source("fn_tidy_mirt_coefs.R")

tidy_2pl <- tidy_mirt_coefs(coefs_2pl) %>% 
  # keep only the a and b parameters from the 2PL model
  filter(par %in% c("a", "b"))

Here is the result:

tidy_2pl %>%
  mutate(ci = str_glue("[{round(CI_2.5, 2)}, {round(CI_97.5, 2)}]")) %>% 
  select(-starts_with("CI_")) %>% 
  pivot_wider(names_from = "par", values_from = c(est, ci), names_glue = "{par}_{.value}") %>% 
  gt() %>% 
  fmt_number(columns = contains("_est"), decimals = 2) %>%
  data_color(
    columns = contains("a_est"),
    colors = scales::col_numeric(palette = c("Greens"), domain = NULL)
  ) %>%
  data_color(
    columns = contains("b_est"),
    colors = scales::col_numeric(palette = c("Blues"), domain = NULL)
  ) %>%
  tab_spanner(label = "Discrimination", columns = contains("a_")) %>%
  tab_spanner(label = "Difficulty", columns = contains("b_")) %>%
  cols_label(
    a_est = "Est.",
    b_est = "Est.",
    a_ci = "95% CI",
    b_ci = "95% CI"
  )
Question Discrimination Difficulty
Est. 95% CI Est. 95% CI
A1 1.22 [1.02, 1.41] −2.95 [-3.3, -2.59]
A2 0.56 [0.48, 0.65] 1.67 [1.41, 1.93]
A3 1.35 [1.24, 1.47] −0.54 [-0.61, -0.46]
A4 1.18 [1.07, 1.29] −0.59 [-0.67, -0.5]
A5 1.47 [1.34, 1.59] 0.08 [0.02, 0.14]
A6 1.56 [1.42, 1.7] −0.86 [-0.94, -0.78]
A7 1.29 [1.17, 1.41] −0.88 [-0.97, -0.8]
A8 1.16 [1.06, 1.27] 0.07 [-0.01, 0.14]
A9 2.02 [1.86, 2.18] 0.13 [0.07, 0.18]
A10 1.77 [1.63, 1.91] −0.14 [-0.2, -0.09]
A11 1.68 [1.55, 1.82] −0.28 [-0.34, -0.22]
A12 0.89 [0.79, 0.99] −1.09 [-1.22, -0.96]
A13 0.72 [0.63, 0.81] 0.85 [0.72, 0.99]
A14 1.44 [1.31, 1.56] −0.26 [-0.32, -0.19]
A15 0.85 [0.76, 0.94] 0.21 [0.12, 0.3]
A16 1.54 [1.39, 1.69] 1.30 [1.19, 1.4]
A17 1.23 [1.12, 1.34] −0.40 [-0.48, -0.33]
A18 1.24 [1.12, 1.35] 0.55 [0.47, 0.63]
A19 1.31 [1.19, 1.43] 0.64 [0.56, 0.72]
A20 2.33 [2.13, 2.53] −0.74 [-0.8, -0.68]
A21 2.14 [1.97, 2.31] −0.05 [-0.1, 0.01]
A22 2.00 [1.84, 2.16] −0.05 [-0.11, 0]
A23 1.88 [1.72, 2.03] 0.27 [0.21, 0.33]
A24 1.44 [1.32, 1.56] −0.31 [-0.38, -0.25]
A25 0.53 [0.45, 0.62] −1.13 [-1.33, -0.92]
A26 2.42 [2.2, 2.65] −1.04 [-1.1, -0.97]
A27 1.99 [1.83, 2.16] 0.35 [0.29, 0.41]
A28 1.85 [1.7, 2] 0.21 [0.15, 0.27]
A29 0.91 [0.82, 1.01] 0.00 [-0.09, 0.09]
A30 1.26 [1.14, 1.38] −1.08 [-1.19, -0.98]
A31 1.25 [1.14, 1.36] 0.54 [0.46, 0.62]
A32 1.60 [1.45, 1.75] 1.15 [1.06, 1.24]
A33 0.94 [0.83, 1.04] −1.51 [-1.67, -1.35]
A34 0.91 [0.81, 1] −0.51 [-0.61, -0.42]
A35 1.28 [1.16, 1.4] 0.73 [0.65, 0.82]
A36 1.25 [1.12, 1.37] 1.28 [1.16, 1.39]
tidy_2pl_wide <- tidy_2pl %>% 
  pivot_wider(names_from = "par", values_from = c(est, CI_2.5, CI_97.5), names_glue = "{par}_{.value}")

wright_plot <- tidy_2pl_wide %>% 
  mutate(qnum = parse_number(Question)) %>% 
  ggplot(aes(
    x = a_est,
    y = b_est
  )) +
  geom_errorbar(aes(ymin = b_CI_2.5, ymax = b_CI_97.5), width = 0, alpha = 0.5) +
  geom_errorbar(aes(xmin = a_CI_2.5, xmax = a_CI_97.5), width = 0, alpha = 0.5) +
  geom_text_repel(aes(label = Question), alpha = 0.5) +
  geom_point() +
  theme_minimal() +
  labs(x = "Discrimination",
       y = "Difficulty")

wright_plot

tidy_2pl_wide %>% 
  write_csv("output/eth_pre_2pl-results.csv")

3.3 Comparing years and classes

Do students from different programmes of study have different distributions of ability?

3.3.1 Differences between years

Compare the distribution of abilities in the year groups (though in this case there is only one).

ggplot(test_scores_with_2pl_ability, aes(F1, fill = as.factor(year), colour = as.factor(year))) +
  geom_density(alpha=0.5) + 
  scale_x_continuous(limits = c(-3.5,3.5)) +
  labs(title = "Density plot", 
       subtitle = "Ability grouped by year of taking the test", 
       x = "Ability", y = "Density",
       fill = "Year", colour = "Year") +
  theme_minimal()

3.3.2 Differences between classes

Compare the distribution of abilities in the various classes.

ggplot(test_scores_with_2pl_ability, aes(x = F1, y = class, colour = class, fill = class)) +
  geom_density_ridges(alpha = 0.5) + 
  scale_x_continuous(limits = c(-3.5,3.5)) +
  guides(fill = FALSE, colour = FALSE) +
  labs(title = "Density plot", 
       subtitle = "Ability grouped by class of taking the test", 
       x = "Ability", y = "Class") +
  theme_minimal()
## Warning: `guides(<scale> = FALSE)` is deprecated. Please use `guides(<scale> =
## "none")` instead.
## Picking joint bandwidth of 0.194

3.4 Information curves

theta <- seq(-6, 6, by=0.05)

info_matrix <- testinfo(fit_2pl, theta, individual = TRUE)
colnames(info_matrix) <- item_info %>% pull(question)
item_info_data <- info_matrix %>% 
  as_tibble() %>% 
  bind_cols(theta = theta) %>% 
  pivot_longer(cols = -theta, names_to = "item", values_to = "info_y") %>% 
  left_join(item_info %>% select(item = question), by = "item") %>% 
  mutate(item = fct_reorder(item, parse_number(item)))

3.4.1 Test information curve

item_info_data %>% 
  group_by(theta) %>% 
  summarise(info_y = sum(info_y)) %>% 
  ggplot(aes(x = theta, y = info_y)) +
  geom_line() +
  labs(x = "Ability", y = "Information", title = "ETH s21t") +
  theme_minimal()

ggsave("output/eth_pre_info.pdf", width = 10, height = 6, units = "cm")

This shows that the information given by the test is focused reasonably centrally around the mean ability level.

3.4.2 Item information curves

Breaking this down by question helps to highlight those questions that are most/least informative:

item_info_data %>% 
  ggplot(aes(x = theta, y = info_y, colour = item)) +
  geom_line() +
  scale_colour_viridis_d(option = "plasma", end = 0.8, direction = -1) +
  facet_wrap(vars(item)) +
  labs(y = "Information") +
  theme_minimal()

3.5 Total information

Using mirt’s areainfo function, we can find the total area under the information curves.

irt_info <- areainfo(fit_2pl, c(-4,4))
irt_info %>% gt()
LowerBound UpperBound Info TotalInfo Proportion nitems
-4 4 49.13213 50.42239 0.9744109 36

This shows that the total information in all items is 50.4223874.

3.5.1 Information by item

tidy_info <- item_info %>%
  mutate(item_num = row_number()) %>% 
  mutate(TotalInfo = purrr::map_dbl(
    item_num,
    ~ areainfo(fit_2pl,
               c(-4, 4),
               which.items = .x) %>% pull(TotalInfo)
  ))

tidy_info %>%
  select(-item_num) %>% 
  arrange(-TotalInfo) %>% 
  #group_by(outcome) %>% 
  gt() %>% 
  fmt_number(columns = contains("a_"), decimals = 2) %>%
  fmt_number(columns = contains("b_"), decimals = 2) %>%
  data_color(
    columns = contains("info"),
    colors = scales::col_numeric(palette = c("Greens"), domain = NULL)
  ) %>%
  data_color(
    columns = contains("outcome"),
    colors = scales::col_factor(palette = c("viridis"), domain = NULL)
  ) %>%
  cols_label(
    TotalInfo = "Information"
  )
question description Information
A26 int(poly) 2.4238253
A20 (exp)' 2.3258216
A21 (ln(sin))' 2.1365919
A9 trig.ineq. 2.0158405
A22 hyp.functions 1.9989907
A27 int(exp) 1.9938718
A23 slope tangent 1.8750694
A28 Int = 0 1.8501832
A10 trig.identity 1.7675098
A11 period 1.6839435
A33 difference vectors 1.6041930
A6 logs 1.5577381
A16 diff.quotient 1.5386847
A5 logs 1.4663312
A24 IVT 1.4390034
A14 limit 1.4350697
A3 arithmetic rules 1.3547187
A19 quotient rule 1.3093576
A7 graph translation 1.2867712
A30 FtoC graph 1.2809342
A31 int(abs) 1.2631005
A32 FtoC algebra 1.2503874
A34 normal vector 1.2462025
A18 product rule 1.2373800
A17 graph f' 1.2268893
A1 linear function 1.2188793
A4 Easy ineq. 1.1804333
A8 sine pi/3 1.1623303
A35 intersect planes 0.9353326
A29 int even funct 0.9128694
A36 parallel planes 0.9058856
A12 rational exponents 0.8887859
A15 series 0.8494321
A13 ellipsoid 0.7169794
A2 3D 0.5552891
A25 velocity 0.5277613

Restricting instead to the range \(-2\leq\theta\leq2\):

tidy_info <- item_info %>%
  mutate(item_num = row_number()) %>% 
  mutate(TotalInfo = purrr::map_dbl(
    item_num,
    ~ areainfo(fit_2pl,
               c(-2, 2),
               which.items = .x) %>% pull(Info)
  ))

tidy_info %>%
  select(-item_num) %>% 
  arrange(-TotalInfo) %>% 
  #group_by(outcome) %>% 
  gt() %>% 
  fmt_number(columns = contains("a_"), decimals = 2) %>%
  fmt_number(columns = contains("b_"), decimals = 2) %>%
  data_color(
    columns = contains("info"),
    colors = scales::col_numeric(palette = c("Greens"), domain = NULL)
  ) %>%
  data_color(
    columns = contains("outcome"),
    colors = scales::col_factor(palette = c("viridis"), domain = NULL)
  ) %>%
  cols_label(
    TotalInfo = "Information"
  )
question description Information
A26 int(poly) 2.2090139
A20 (exp)' 2.2045246
A21 (ln(sin))' 2.0775693
A9 trig.ineq. 1.9433205
A22 hyp.functions 1.9265566
A27 int(exp) 1.9039799
A23 slope tangent 1.7785873
A28 Int = 0 1.7545283
A10 trig.identity 1.6643308
A11 period 1.5605291
A5 logs 1.3171858
A6 logs 1.3144395
A24 IVT 1.2724263
A14 limit 1.2722619
A33 difference vectors 1.2662362
A3 arithmetic rules 1.1482188
A16 diff.quotient 1.1403300
A19 quotient rule 1.0806177
A30 FtoC graph 1.0322320
A32 FtoC algebra 1.0269435
A17 graph f' 1.0141086
A18 product rule 1.0104220
A7 graph translation 1.0086145
A8 sine pi/3 0.9547704
A4 Easy ineq. 0.9399399
A31 int(abs) 0.9357119
A34 normal vector 0.8652332
A29 int even funct 0.6597833
A36 parallel planes 0.6349100
A15 series 0.5847449
A12 rational exponents 0.5619252
A35 intersect planes 0.5401834
A13 ellipsoid 0.4171787
A1 linear function 0.2895158
A25 velocity 0.2433460
A2 3D 0.2430917

3.6 Response curves

These show the probability of a correct response at each ability level:

trace_data <- probtrace(fit_2pl, theta) %>% 
  as_tibble() %>% 
  bind_cols(theta = theta) %>% 
  pivot_longer(cols = -theta, names_to = "level", values_to = "y") %>% 
  separate(level, into = c("item", NA, "score"), sep = "\\.") %>% 
  mutate(item = fct_reorder(item, parse_number(item))) %>% 
  filter(score == 1)

trace_data %>% 
  ggplot(aes(x = theta, y = y, colour = score)) +
  geom_line() +
  scale_colour_viridis_d(option = "plasma", end = 0.8, direction = -1, guide = "none") +
  facet_wrap(vars(item)) +
  labs(y = "Probability of correct response") +
  theme_minimal()

These can all be overlaid on a single plot:

plt <- trace_data %>% 
  ggplot(aes(x = theta, y = y, colour = item, text = item)) +
  geom_line() +
  scale_colour_viridis_d(option = "plasma", end = 0.8, direction = -1) +
  labs(y = "Probability of correct response") +
  theme_minimal()

ggplotly(plt, tooltip = "text")
ggsave(plot = plt, file = "output/eth_pre_iccs-superimposed.pdf", width = 20, height = 14, units = "cm")

Here we highlight some items to illustrate the role of the difficulty and discrimination parameters

highlight_items <- c("A1", "A2", "A26", "A27")
item_colours <- c(viridis::viridis_pal(option = "plasma", end = 0.8, direction = -1)(length(highlight_items)), "black")

wright_plot_shaded <- tidy_2pl_wide %>%
  mutate(item = Question) %>%
  mutate(
    item_colour = case_when(item %in% highlight_items ~ item,
                            TRUE ~ "ignore"),
    item_size = if_else(item %in% highlight_items, 4, 3)
  ) %>% 
  ggplot(aes(
    x = a_est,
    y = b_est,
    colour = item_colour
  )) +
  geom_errorbar(aes(ymin = b_CI_2.5, ymax = b_CI_97.5), width = 0, alpha = 0.5) +
  geom_errorbar(aes(xmin = a_CI_2.5, xmax = a_CI_97.5), width = 0, alpha = 0.5) +
  geom_text_repel(aes(label = Question, size = item_size), alpha = 0.5) +
  geom_point() +
  scale_colour_manual("item", values = item_colours, limits = highlight_items, guide = "none") +
  scale_size(guide = "none", range = c(2.5, 4.5)) +
  theme_minimal() +
  coord_flip() +
  labs(x = "Discrimination",
       y = "Difficulty")

trace_examples <- trace_data %>% 
  filter(item %in% highlight_items) %>% 
  ggplot(aes(x = theta, y = y, colour = item)) +
  geom_line() +
  # ggrepel::geom_label_repel(
  #   data = trace_data %>% filter(item %in% highlight_items, y > 0.5) %>% group_by(item) %>% slice_min(y, n = 1),
  #   aes(label = item),
  #   box.padding = 0.5,
  #   min.segment.length = Inf,
  #   show.legend = FALSE
  # ) +
  geom_label(
    data = trace_data %>% filter(item %in% c("A1", "A27"), y > 0.45) %>% group_by(item) %>% slice_min(y, n = 1),
    aes(label = item),
    show.legend = FALSE
  ) +
  geom_label(
    data = trace_data %>% filter(item %in% c("A2", "A26"), y > 0.55) %>% group_by(item) %>% slice_min(y, n = 1),
    aes(label = item),
    show.legend = FALSE
  ) +
  scale_colour_viridis_d("item", option = "plasma", end = 0.8, direction = -1, guide = "none") +
  labs(x = "Ability", y = "Probability of correct response") +
  theme_minimal()

combined_plot <- patchwork::wrap_plots(wright_plot_shaded, trace_examples)
ggsave("output/eth_pre_wrightmap_highlight.pdf", plot = combined_plot, units = "cm", width = 18, height = 9)
combined_plot

removed_items <- read_csv("data-eth/eth-metadata.csv") %>% filter(is.na(post)) %>% select(pre) %>% deframe()
highlight_removed_items <- trace_data %>% 
  mutate(highlight_item = item %in% removed_items) %>% 
  mutate(line_width = ifelse(highlight_item, 1, 0.5))

highlight_removed_items %>% 
  ggplot(aes(x = theta, y = y, colour = item, text = item, alpha = highlight_item)) +
  geom_line(aes(size = highlight_item)) +
  #geom_point(data = highlight_removed_items %>% filter(highlight_item == TRUE, theta == 0)) +
  ggrepel::geom_label_repel(
    #data = highlight_removed_items %>% filter(highlight_item == TRUE) %>% group_by(item) %>% mutate(item_num = cur_group_id()) %>% ungroup() %>% mutate(threshold = seq(5, -5, len=max(item_num))[item_num]) %>% filter(theta >= threshold) %>% group_by(item_num) %>% slice(n=1),
    #data = highlight_removed_items %>% filter(highlight_item == TRUE, y >= 0.5) %>% group_by(item) %>% slice_min(y, n = 1),
    data = highlight_removed_items %>% filter(highlight_item == TRUE, theta == 0),
    aes(label = item),
    box.padding = 0.5,
    max.overlaps = Inf,
    min.segment.length = 0,
    show.legend = FALSE
  ) +
  scale_colour_viridis_d("Question", option = "plasma", end = 0.8, direction = -1) +
  scale_size_manual(values = c("FALSE" = 0.6, "TRUE" = 0.9), guide = "none") +
  scale_alpha_discrete(guide = "none", range = c(0.2, 1)) +
  labs(x = "Ability", y = "Expected score") +
  theme_minimal() +
  theme(legend.position="bottom",#legend.title=element_blank(),
      legend.margin = margin(0, 0, 0, 0),
      legend.spacing.x = unit(1, "mm"),
      legend.spacing.y = unit(0, "mm")) +
  guides(colour = guide_legend(nrow = 4))

ggsave(file = "output/eth_pre_iccs-highlight.pdf", width = 16, height = 12, units = "cm")

4 Checking 3PL

The 3 parameter IRT model adds a third “guessing” parameter for each item, which gives the lower asymptote of the response curve.

Given that the test offered an “I don’t know” option (which we have mapped to a score of 0), we would expect that students are less likely to be guessing when answering these multiple-choice questions. So, we would expect to see the lower aymptotes being closer to 0 than would ordinarily be expected for a 4- or 5-option MCQ.

fit_3pl <- mirt(
  data = item_scores, # just the columns with question scores
  model = 1,          # number of factors to extract
  itemtype = "3PL",   # 3 parameter logistic model
  SE = TRUE           # estimate standard errors
  )

residuals_3pl <- residuals(fit_3pl, type = "Q3") %>% data.frame

coefs_3pl <- coef(fit_3pl, IRTpars = TRUE)

test_scores_with_3pl_ability <- test_scores %>%
  bind_cols(fscores(fit_3pl, method = "EAP"))

4.1 Local independence

We compute Yen’s \(Q_3\) (1984) to check for any dependence between items after controlling for \(\theta\). This gives a score for each pair of items, with scores above 0.2 regarded as problematic (see DeMars, p. 48).

residuals_3pl  %>% as.matrix() %>% 
  corrplot::corrplot(type = "upper")

This shows that most item pairs are independent, with only a couple of pairs showing cause for concern:

residuals_3pl %>%
  rownames_to_column(var = "q1") %>%
  as_tibble() %>% 
  pivot_longer(cols = starts_with("A"), names_to = "q2", values_to = "Q3_score") %>% 
  filter(abs(Q3_score) > 0.2) %>% 
  filter(parse_number(q1) < parse_number(q2)) %>%
  gt()
q1 q2 Q3_score
A18 A19 0.6745194
A34 A35 0.2115169
  • Items A18 and A19 are on the product and quotient rules, with an identical structure in each case (giving all the values that need to be used in the relevant formula).

  • Items A34 and A35 are are both about planes in vector geometry.

Given that this violation of the local independence assumption is very mild, we proceed using this model.

4.2 Model parameters

tidy_3pl <- tidy_mirt_coefs(coefs_3pl) %>% 
  filter(par %in% c("a", "b", "g"))

Here is a nicely formatted table of the result:

tidy_3pl %>%
  mutate(ci = str_glue("[{round(CI_2.5, 2)}, {round(CI_97.5, 2)}]")) %>% 
  select(-starts_with("CI_")) %>% 
  pivot_wider(names_from = "par", values_from = c(est, ci), names_glue = "{par}_{.value}") %>% 
  gt() %>% 
  fmt_number(columns = contains("_est"), decimals = 2) %>%
  data_color(
    columns = contains("a_est"),
    colors = scales::col_numeric(palette = c("Greens"), domain = NULL)
  ) %>%
  data_color(
    columns = contains("b_est"),
    colors = scales::col_numeric(palette = c("Blues"), domain = NULL)
  ) %>%
  data_color(
    columns = contains("g_est"),
    colors = scales::col_numeric(palette = c("Reds"), domain = NULL)
  ) %>%
  tab_spanner(label = "Discrimination", columns = contains("a_")) %>%
  tab_spanner(label = "Difficulty", columns = contains("b_")) %>%
  tab_spanner(label = "Guessing", columns = contains("g_")) %>%
  cols_label(
    a_est = "Est.",
    b_est = "Est.",
    g_est = "Est.",
    a_ci = "95% CI",
    b_ci = "95% CI",
    g_ci = "95% CI"
  )
Question Discrimination Difficulty Guessing
Est. 95% CI Est. 95% CI Est. 95% CI
A1 1.18 [0.99, 1.37] −3.01 [-3.44, -2.58] 0.02 [-0.17, 0.21]
A2 0.83 [0.51, 1.14] 1.81 [1.55, 2.07] 0.10 [0.03, 0.18]
A3 1.66 [1.38, 1.94] −0.23 [-0.42, -0.05] 0.15 [0.06, 0.24]
A4 1.38 [1.14, 1.62] −0.30 [-0.55, -0.06] 0.13 [0.02, 0.24]
A5 1.74 [1.49, 2] 0.22 [0.11, 0.33] 0.07 [0.02, 0.12]
A6 1.66 [1.4, 1.92] −0.72 [-0.94, -0.49] 0.08 [-0.04, 0.21]
A7 1.29 [1.12, 1.47] −0.86 [-1.11, -0.62] 0.01 [-0.11, 0.13]
A8 1.19 [1.01, 1.37] 0.09 [-0.07, 0.26] 0.01 [-0.06, 0.08]
A9 2.04 [1.82, 2.26] 0.14 [0.07, 0.21] 0.00 [-0.02, 0.03]
A10 2.08 [1.81, 2.35] −0.01 [-0.1, 0.09] 0.07 [0.03, 0.12]
A11 1.90 [1.64, 2.15] −0.15 [-0.27, -0.03] 0.07 [0.01, 0.13]
A12 0.96 [0.7, 1.22] −0.81 [-1.58, -0.04] 0.11 [-0.18, 0.41]
A13 0.88 [0.63, 1.13] 1.04 [0.79, 1.3] 0.08 [-0.01, 0.17]
A14 1.64 [1.4, 1.88] −0.09 [-0.23, 0.05] 0.08 [0.01, 0.15]
A15 0.98 [0.72, 1.24] 0.42 [0.08, 0.76] 0.08 [-0.05, 0.2]
A16 1.73 [1.45, 2] 1.28 [1.18, 1.37] 0.01 [0, 0.03]
A17 1.23 [1.12, 1.34] −0.40 [-0.47, -0.32] 0.00 [-0.01, 0.01]
A18 1.25 [1.14, 1.37] 0.55 [0.47, 0.63] 0.00 [0, 0]
A19 1.33 [1.21, 1.45] 0.64 [0.56, 0.72] 0.00 [0, 0]
A20 2.32 [2.02, 2.62] −0.72 [-0.84, -0.6] 0.01 [-0.07, 0.08]
A21 2.15 [1.98, 2.32] −0.03 [-0.09, 0.02] 0.00 [0, 0]
A22 2.07 [1.86, 2.27] −0.02 [-0.09, 0.05] 0.01 [-0.01, 0.04]
A23 1.89 [1.74, 2.05] 0.28 [0.22, 0.33] 0.00 [0, 0]
A24 1.44 [1.32, 1.57] −0.30 [-0.37, -0.24] 0.00 [-0.01, 0.01]
A25 0.54 [0.45, 0.62] −1.11 [-1.36, -0.85] 0.00 [-0.04, 0.05]
A26 2.39 [2.16, 2.61] −1.03 [-1.11, -0.96] 0.00 [-0.01, 0.01]
A27 2.02 [1.85, 2.18] 0.36 [0.3, 0.41] 0.00 [0, 0]
A28 1.87 [1.72, 2.02] 0.22 [0.16, 0.28] 0.00 [0, 0.01]
A29 0.92 [0.83, 1.01] 0.01 [-0.08, 0.09] 0.00 [0, 0.01]
A30 1.25 [1.12, 1.38] −1.08 [-1.21, -0.95] 0.00 [-0.04, 0.05]
A31 1.44 [1.2, 1.68] 0.62 [0.51, 0.74] 0.04 [0, 0.09]
A32 1.63 [1.48, 1.79] 1.14 [1.05, 1.23] 0.00 [0, 0]
A33 0.93 [0.81, 1.04] −1.50 [-1.78, -1.21] 0.01 [-0.1, 0.12]
A34 0.91 [0.81, 1.01] −0.50 [-0.63, -0.37] 0.00 [-0.03, 0.04]
A35 1.31 [1.15, 1.46] 0.74 [0.64, 0.83] 0.00 [-0.02, 0.03]
A36 1.64 [1.32, 1.97] 1.28 [1.17, 1.38] 0.04 [0.02, 0.07]
tidy_3pl_wide <- tidy_3pl %>% 
  pivot_wider(names_from = "par", values_from = c(est, CI_2.5, CI_97.5), names_glue = "{par}_{.value}")

tidy_3pl_wide %>% 
  mutate(qnum = parse_number(Question)) %>% 
  ggplot(aes(
    x = qnum,
    y = g_est
  )) +
  geom_errorbar(aes(ymin = g_CI_2.5, ymax = g_CI_97.5), width = 0, alpha = 0.5) +
  geom_text_repel(aes(label = Question), alpha = 0.5) +
  geom_point() +
  theme_minimal() +
  labs(x = "Question",
       y = "Guessing parameter")

tidy_3pl_wide %>% 
  write_csv("output/eth_pre_3pl-results.csv")

This shows that overall guessing appears to be rare; certainly below the level that might be expected if students of very low ability were simply choosing an answer at random.

About this report

This report supports the analysis in the following paper:

[citation needed]

Packages

In this analysis we used the following packages. You can learn more about each one by clicking on the links below.

  • mirt: For IRT analysis
  • psych: For factor analysis
  • tidyverse: For data wrangling and visualisation
  • reshape: For reshaping nested lists
  • vroom: For reading in many files at once
  • broom: For tidying model output
  • fs: For file system operations
  • gt: For formatting tables
  • knitr: For markdown tables
  • ggrepel: For labelling points without overlap
  • skimr: For data frame level summary
  • ggridges: For ridge plots